I've been going through MIT OCW's Linear Algebra lectures, and Professor Strang demonstrated the algorithmic steps to solving the equation $Ax = 0$ to find the nullspace of $A$.
At around nineteen minutes in, he starts introducing pivot and free columns of an echelon matrix:
$$U = \begin{bmatrix}1&2&2&4\\0&0&2&4\\0&0&0&0\end{bmatrix}$$
Where the second and forth columns represent free columns and the first and third pivot columns. I understand pivot columns, but then he goes onto say that the free columns correspond to free variables in the solution $x$.
Finally, he sets $x_2 = 1$ and $x_4 = 0$ and solves for that system, calling it a special solution. He then switches them around with $x_2 = 0$ and $x_4 = 1$, concluding with the solution set of:
$$x = c\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + d\begin{bmatrix}0\\0\\-2\\1\end{bmatrix}$$
I want to know why these variables are essentially "free" and are able to be whatever. My current understanding is that $1,0$ and $0,1$ are substituted because they are somehow "basis" vectors in which the linear combination would give us all solutions. But what makes these specific columns "free"? In what way are they special, giving them the ability to take any value?
I think the situation is clearest if you perform the full row reduction first; then your system is equivalent to the one given by the matrix $$\begin{pmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 2 \end{pmatrix}.$$
This shows that the pivots are independent of one another, yet that they depend on the so-called free columns. Explicitly, the system is given by $x_1 + 2x_2 = 0$, $x_3 + 2x_4= 0$, and the dependence may be rewritten as $x_1 = -2x_2$, $x_3 = -2x_4$. Now it is transparent that $x_2,x_4$ may take arbitrary values, and this forces the pivots to take specific values given by the dependence relations. This generalizes to any homogeneous system.